$\overline{AB}$ = $\sqrt{61}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $\sqrt{61}$ $?$ $ \sin( \angle BAC ) = \frac{5\sqrt{61} }{61}, \cos( \angle BAC ) = \frac{6\sqrt{61} }{61}, \tan( \angle BAC ) = \dfrac{5}{6}$
$\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{\sqrt{61}} $ $ \overline{BC}=\sqrt{61} \cdot \sin( \angle BAC ) = \sqrt{61} \cdot \frac{5\sqrt{61} }{61} = 5$